Is ${481680}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {481680}= &&{4}\cdot100000+ \\&&{8}\cdot10000+ \\&&{1}\cdot1000+ \\&&{6}\cdot100+ \\&&{8}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {481680}= &&{4}(99999+1)+ \\&&{8}(9999+1)+ \\&&{1}(999+1)+ \\&&{6}(99+1)+ \\&&{8}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {481680}= &&\gray{4\cdot99999}+ \\&&\gray{8\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {4}+{8}+{1}+{6}+{8}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${481680}$ is divisible by $9$ if ${ 4}+{8}+{1}+{6}+{8}+{0}$ is divisible by $9$ Add the digits of ${481680}$ $ {4}+{8}+{1}+{6}+{8}+{0} = {27} $ If ${27}$ is divisible by $9$ , then ${481680}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${481680}$ must also be divisible by $9$.